Problem: $ B = \left[\begin{array}{rr}3 & 2 \\ 1 & -1\end{array}\right]$ $ E = \left[\begin{array}{rr}3 & 1 \\ -2 & 5\end{array}\right]$ What is $ B E$ ?
Answer: Because $ B$ has dimensions $(2\times2)$ and $ E$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ B E = \left[\begin{array}{rr}{3} & {2} \\ {1} & {-1}\end{array}\right] \left[\begin{array}{rr}{3} & \color{#DF0030}{1} \\ {-2} & \color{#DF0030}{5}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ B$ , with the corresponding elements in column $j$ of the second matrix, $ E$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ B$ with the first element in ${\text{column }1}$ of $ E$ , then multiply the second element in ${\text{row }1}$ of $ B$ with the second element in ${\text{column }1}$ of $ E$ , and so on. Add the products together. $ \left[\begin{array}{rr}{3}\cdot{3}+{2}\cdot{-2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ B$ with the corresponding elements in ${\text{column }1}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{3}\cdot{3}+{2}\cdot{-2} & ? \\ {1}\cdot{3}+{-1}\cdot{-2} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ B$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{3}\cdot{3}+{2}\cdot{-2} & {3}\cdot\color{#DF0030}{1}+{2}\cdot\color{#DF0030}{5} \\ {1}\cdot{3}+{-1}\cdot{-2} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{3}\cdot{3}+{2}\cdot{-2} & {3}\cdot\color{#DF0030}{1}+{2}\cdot\color{#DF0030}{5} \\ {1}\cdot{3}+{-1}\cdot{-2} & {1}\cdot\color{#DF0030}{1}+{-1}\cdot\color{#DF0030}{5}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}5 & 13 \\ 5 & -4\end{array}\right] $